Divide the following complex numbers. $ \dfrac{-2+i}{1+2i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${1-2i}$ $ \dfrac{-2+i}{1+2i} = \dfrac{-2+i}{1+2i} \cdot \dfrac{{1-2i}}{{1-2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-2+i) \cdot (1-2i)} {(1+2i) \cdot (1-2i)} = \dfrac{(-2+i) \cdot (1-2i)} {1^2 - (2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-2+i) \cdot (1-2i)} {(1)^2 - (2i)^2} = $ $ \dfrac{(-2+i) \cdot (1-2i)} {1 + 4} = $ $ \dfrac{(-2+i) \cdot (1-2i)} {5} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-2+i}) \cdot ({1-2i})} {5} = $ $ \dfrac{{-2} \cdot {1} + {1} \cdot {1 i} + {-2} \cdot {-2 i} + {1} \cdot {-2 i^2}} {5} $ Evaluate each product of two numbers. $ \dfrac{-2 + 1i + 4i - 2 i^2} {5} $ Finally, simplify the fraction. $ \dfrac{-2 + 1i + 4i + 2} {5} = \dfrac{0 + 5i} {5} = i $